24- What Is Disjunctive Addition And Conjunctive Simplification In Rules Of Inference In Logic

Описание к видео 24- What Is Disjunctive Addition And Conjunctive Simplification In Rules Of Inference In Logic

This Tutorials are on discrete mathematics for computer science and Information technology students of Btech, BSc., MSc, Mtech and others studying in universities as well. Rules of inference are no more than valid arguments. The simplest yet most fundamental valid arguments are

modus ponens: p q, p, q

modus tollens: p q, q, p
Latin phrases modus ponens and modus tollens carry the meaning of ``method of affirming'' and ``method of denying'' respectively. That they are valid can be easily established. Modus tollens, for instance, can be seen or derived by the following truth table

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shows the validity of the argument form.


disjunctive addition: p, p q.

conjunctive addition: p, q, p q.

conjunctive simplification: p q, p

disjunctive syllogism: p q, q, p

hypothetical syllogism: p q, q r, p r

division into cases: p q, p r, q r, r

rule of contradiction: p contradiction, p
The validity of the above argument forms can all be easily verified via truth tables. In fact the case of ``division into cases'' has been proven in example 2. These rules may not mathematically look very familiar. But it is most likely that everyone has used them all, individually or jointly, at some stage subconsciously.
Examples

Show that the following argument form
p q, q r, p s t, r, q u s, t (*)
is valid by breaking it into a list of known elementary valid argument forms or rules.
Solution We'll treat all the rules of inference introduced earlier in this subsection as the known elementary argument forms. The logical inference for the argument form in the question is as follows.
(1) q r premise
r premise
q by modus tollens
(2) p q premise
q by (1)
p by disjunctive syllogism
(3) q u s premise
q by (1)
u s by modus ponens
(4) u s by (3)
s by conjunctive simplification
(5) p by (2)
s by (4)
p s by conjunctive addition
(6) p s t premise
p s by (5)
t by modus ponens
Can you explain, in additional details, how those 6 proof steps in the above example come into existence?

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